Answer
$\frac{1}{x}+\frac{1}{x^2+1}-\frac{x+2}{(x^2+1)^2}$
Work Step by Step
Step 1. The denominator can not be further factorized, assume the end results as: $\frac{A}{x}+\frac{Bx+C}{x^2+1}+\frac{Dx+E}{(x^2+1)^2}$
Step 2. Combine the above functions as:
$\frac{A(x^2+1)^2+x(Bx+c)(x^2+1)+x(Dx+E)}{x(x^2+1)^2}=\frac{(A+B)x^4+(C)x^3+(2A+B+D)x^2+(C+E)x+(A)}{x(x^2+1)^2}$
Step 3. Compare the above result with the original expression to set up the following system of equations:
\begin{cases} A+B=1 \\ C=1\\2A+B+D=1\\C+E=-1\\A=1 \end{cases}
Step 4. Use substitution to solve the above equations and get $A=1, B=0,C=1,D=-1,E=-2$
Step 5. Write the final results as:
$\frac{1}{x}+\frac{1}{x^2+1}-\frac{x+2}{(x^2+1)^2}$
